8 Let \(I = \int _ { 2 } ^ { 5 } \frac { 5 } { x + \sqrt { } ( 6 - x ) } \mathrm { d } x\).
- Using the substitution \(u = \sqrt { } ( 6 - x )\), show that
$$I = \int _ { 1 } ^ { 2 } \frac { 10 u } { ( 3 - u ) ( 2 + u ) } \mathrm { d } u$$
- Hence show that \(I = 2 \ln \left( \frac { 9 } { 2 } \right)\).